Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{-2k + 10}{k^3 - 6k^2 + 5k} \times \dfrac{3k^3 + 3k^2 - 6k}{3k + 24} $
Solution: First factor out any common factors. $z = \dfrac{-2(k - 5)}{k(k^2 - 6k + 5)} \times \dfrac{3k(k^2 + k - 2)}{3(k + 8)} $ Then factor the quadratic expressions. $z = \dfrac {-2(k - 5)} {k(k - 1)(k - 5)} \times \dfrac {3k(k - 1)(k + 2)} {3(k + 8)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(k - 5) \times 3k(k - 1)(k + 2) } { k(k - 1)(k - 5) \times 3(k + 8)} $ $z = \dfrac {-6k(k - 1)(k + 2)(k - 5)} {3k(k - 1)(k - 5)(k + 8)} $ Notice that $(k - 1)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-6k\cancel{(k - 1)}(k + 2)(k - 5)} {3k\cancel{(k - 1)}(k - 5)(k + 8)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $z = \dfrac {-6k\cancel{(k - 1)}(k + 2)\cancel{(k - 5)}} {3k\cancel{(k - 1)}\cancel{(k - 5)}(k + 8)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $z = \dfrac {-6k(k + 2)} {3k(k + 8)} $ $ z = \dfrac{-2(k + 2)}{k + 8}; k \neq 1; k \neq 5 $